If you used the 8900 kilojoules you expend in energy in one day to heat 46000 {\rm g} of water at 25^\circ {\rm C}, what would be the rise in temperature?
q = mass water x specific heat water x delta T. Solve for delta T. Add 25 if you want to know the final T.
0.015
To calculate the rise in temperature, we can use the equation:
Q = mcΔT
Where:
Q is the amount of heat energy transferred,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.
First, let's determine the variables:
- The energy expended (Q) is given as 8900 kilojoules, which we'll convert to joules by multiplying by 1000: Q = 8900 kilojoules x 1000 = 8,900,000 joules.
- The mass (m) of the water is given as 46000 grams, which we'll convert to kilograms by dividing by 1000: m = 46000 grams / 1000 = 46 kilograms.
- The specific heat capacity (c) of water is approximately 4.18 joules per gram-degree Celsius.
Now, let's plug in the values to calculate the change in temperature (ΔT):
8,900,000 joules = (46 kilograms) x (4.18 joules/(gram-degree Celsius)) x ΔT
Simplifying the equation, we can cancel out the units:
8,900,000 = (46) x (4.18) x ΔT
Now we can solve for ΔT:
ΔT = 8,900,000 / (46 x 4.18)
Using a calculator, we find:
ΔT ≈ 4.89 degrees Celsius
Therefore, the rise in temperature of the 46,000 grams of water would be approximately 4.89 degrees Celsius.